Chapter Six Energy Relationships in Chemical Reactions 06 notes F12.pdf · Thermal energy Energy associated with temperature Kinetic energy: molecular movement Chemical energy ...

1

Chapter Six

Energy Relationships in

Chemical Reactions

2

Energy (U): Capacity to Do WorkRadiant energy

Energy from the sun

Nuclear energyEnergy stored in the nucleus of an atom

Thermal energy Energy associated with temperatureKinetic energy: molecular movement

Chemical energyEnergy stored in chemical bondsPotential energy: Object postion

3

ThermochemistryStudy of heat change in chemical reactionsSystem: The part of the universe being studied

Open: Energy & matter exchange with surroundingClosed: Only energy exchange with surroundingsIsolated: No energy or matter exchange- rare

Surroundings: Part of the universe not being studied

4

Heat Transfer in a System

Heat, q: Transfer of energy due to temperature difference

Exothermic Reaction: System gives off heat (Exiting)Ex: Methane Burning: CH4 + O2 → CO2 + H2OBonds stronger in CO2 and H2O molecules than in CH4 + O2Heat goes from system to surroundings

Endothermic Reaction: System gains heat (Entering)Ex: Ice Melting: H2O (s) → H2O (l)Energy needed to break attractions between H2O moleculesHeat goes from surroundings into the system

5

ThermodynamicsStudy of the conversions between

heat & energy

6

Using ThermodynamicsState of a system:

Describes specific conditions defined by state functionsMeasurement of these propertiesPotential energy of ball at the top of the mountain all

State functions: Properties defined by final - initial values onlyComposition, temperature, pressure, energy, volumeHeight of mountain

Path functions: final - initial values vary by process

Heat, work: interdependent variables# steps taken, Time to top

energy per step, etc.

7

First Law of ThermodynamicsEnergy can be converted from 1 form to another but

cannot be created or destroyedUsystem= Ufinal-Uinitial

Internal Energy of System: Usystem = Ukinetic+ Upotential

Kinetic energy: Amount of molecular motionAssociated with temperature

Potential energy: Energy stored in bondsStrength of chemical bonds

Cannot separate two types of energy For chemical reactions: Ureaction = Uproducts- Ureactants

8

Work and HeatEnergy transfer is caused by changes in heat & work in a system

Usystem = q + wWork (w) = force x distance = -PV

Negative work (-w): System loses energy to surroundingsPositive work (+w): System gains energy from surroundings

Work (w) = -PVV = Vf -Vi

For this situation:+V-w

9

Work and Heat as Path FunctionsA gas expands in volume from 2L to 4L at STP. 350J of energy

are needed to maintain the new conditions. Usys = Ufinal- Uinitial =350J = q + wV = 2L P = 1atm T= 273K

a. Calculate work done against a vacuum & a pressure of 1atm.w = -PV = -(0atm x 2L) = 0atm L = 0Jw = -PV = -(1atm x 2L) = -2atm L = -202.6J (1Latm= 101.32J)

b. Calculate the heat required for each system. q = Usys - w Under a vacuum: q = 350J-0J = 350J (All energy is heat)Under 2.0atm: q = 350J-(-202.6J) = 552.6J (need extra heat!)

10

Enthalpy of

Chemical Reactions

11

Pressure/Volume Relationships with U For systems at constant pressure (usually ~1atm):

U = q + w = qp - PV

Some work used to expand system against surroundings.

Enthalpy: (H) Experimentally measured and tabulatedAt constant pressure: H= U + PV Assume negligible volume change: H= U = qp

There will be slight differences between H and U

12

Real Differences between U and H H measured experimentally as heat

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) H=-367.5kJ/mol

Volume change of H2 gasV=nRT/PT= 298K, P=1.0atm, n=1.0mol H2V= 24.5L

U = H- PVH = -367.5kJPV = 1.0atm x 24.5L = 24.5Latm x 0.101kJ/Latm = 2.5kJU = -367.5kJ -2.5kJ= 370kJ

Difference between H and U is only 2.5kJ~0.7% difference considered negligible

13

Enthalpy of Reaction (H = qrxn)Amount of heat exchanged between system and

surroundings during a chemical reactionHreaction = Hproducts- Hreactants

Exothermic ReactionsTemperature increases in an isolated system.Heat released to surroundings is treated as a product2SO2 (g) + O2(g) 2SO3 (g) + 197.8kJ H = -197.8kJ

Endothermic ReactionsTemperature decreases in an isolated system.Heat absorbed by system is treated as a reactant 197.8kJ + 2SO32SO2 (g) + O2(g) H = +197.8kJ

State function: Independent of path

14

Rules for Working with Heat and Enthalpy1. Use same multiplier as used in chemical reaction

2SO2 (g) + O2(g) 2SO3 (g) H = -197.8kJ

Decrease or increase reactants and products, do the same for H

1SO2 (g) + 1/2O2(g) 1SO3 (g) H = -197.8kJ/2= -98.9kJ

Note: Write H with units that match mole ratio in equationH = -98.9 kJ /mol SO3(g)

2. Reverse the reaction, reverse sign of H2SO3 2SO2 (g) + O2(g) H = +197.8kJ

Use rules to calculate the H for any amount of product or reactant

15

Calculate how much heat is required to decompose15.0g NO2(g) according to this reaction:

2NO(g) + O2(g) 2NO2(g) H = -114 kJHeat required for following reaction:

2NO2(g) 2NO(g) + O2(g) H = +114 kJH = +114 kJ is needed to decompose of 2 moles NO2

Calculate molar mass of NO2 to convert mass to moles of NO2.

Find the amount of heat for the amount of NO2 actually present.

2

2

22

2326.0

0.461

10.15

NONO

NONONO mol

gmol

xg

moles

kJmol

xmol

kJH NO

NOrxn 6.18

1326.0

2114 2

2

16

Calorimetry

17

CalorimetryCalorimetry: Measurement of heat changesCalorimeter:

Device to measure heat (T) produced by a chemical reaction T= Tfinal –Tinitial

Units: C

Specific Heat (s): Heat needed to raise T of 1 gram by 1C. s = q/(m x T) swater = 4.184 J/gCUnits: J/gC

Heat capacity (C): Quantity of heat needed to raise T by 1CC = m x sUnits: J/C or J/K

Heat of Reaction (qrxn) = m x s x T Units: J

18

Commonly Used Specific Heats

19

Calorimetry CalculationsReaction (system) heat is transferred to water (surroundings)

qsys = -qH2O

1. Measure T of waterqH2O = mH2O x sH2O x TH2O

Use q H2O to find system variablesCrossover: qH2O = -qsysqrxn= msys x ssys x Tsys

Always have 2 sets of variables!1 for water (surroundings)1 for reaction (system)

Make 2 columns of variablesSystem: s, m, Tinitial, TfinalWater: s, m, Tinitial, Tfinal

20

A 30.0g sample of metal is heated to 100.0C by placing it in a boiling water bath. The metal is removed from the boiling water, and placed in 25.0g of water (4.18 J/gC) at 22.0C. The final

temp was 27.6C, what is the specific heat of the metal?What is your equation?

What is qmetal?

What is the metal mass?

What is Tmetal ?

What is S. H. metal?

Smetal = qmetal/mTqmetal = - qwaterqwater = masswater x Swater x TwaterTwater = 27.6C – 22.0C = 5.6C qwater = 25.0g x 4.18 J/gC x 5.6C = 585 Jqmetal = - qwater = -585 J 30.0g

(27.6C -100.0C) = -72.4C

CgJ

Cx

gxJSHmetal

27.04.72

10.30

11

585

21

Reaction conducted in a Styrofoam cup calorimeterThe calorimeter does not absorb or give off heat. Heat absorbed by the solution in the calorimeter, qcal is the heat that has been given off by the chemical reaction, -qrxn

Two solutions are mixed in the calorimeter40.0 mL of 1.00 M KOH(aq) 40.0mL of 0.500M H2S04(aq)Ti of both solutions = 21.00C

Data on final solutionDensity: 1.02 g/mL Volume: 80.0 mLS.H.: 4.00 J/gC Temperature: 27.85C

Heat Evolved During a Chemical Reaction

22

Calculate the enthalpy change, H, of this reaction per mole of H2O formed.

What is the chemical reaction? 2 KOH + H2SO4 K2SO4(aq) + 2H2OWhat is the equation for H? Hrxn= qrxn = -qH2OCalculate Heat of Reaction (q). q = S x m x T

s = 4.00 J/gC (assume water)m = 81.6g (mass of both solns) m = 1.02 g/mLx 80.0 mL = 81.6gT = 6.85C (+ temp went up) T = 27.85C - 21.00C = 6.85C

qH2O= 4.00 J/gC x 81.6g x 6.85C = 2236 J

qrxn = -qH2O= -2236 J

Note: this is heat that was transferred into the water IN CALORIMETER, not water in chemical reaction!!!

Amount of heat generated for 81.6g of reactants, not per mole H2O!

23

Calculating the H per mole H2OHow many moles of water are produced in the reaction?

2 KOH + H2SO4 K2SO4(aq) + 2H2O40.0 mL of 1.00 M KOH (aq) = 0.040 mol KOH = 0.040 mol H2O40.0 mL of 0.500 M H2S04(aq) = 0.020 mol H2S04 = 0.040 mol H2O

Calculate H per 1 mole of water produced.Hrxn = qrxn = -2236 JH per mole water = -2236 J / 0.040 mol H2O = -55900 J/molH per mole water = -55.9kJ/mol

24

Heats of Reactions

25

Standard Enthalpy of Formation and Reaction

26

The Direct MethodStandard Enthalpies of Formation, Hf

The standard state of an element: Hf=0Pure element in its most stable form at a pressure of 1 atm and 20CFor solvents in aqueous solution, they are at a concentration of 1 M

H2(g) N2(g) O2(g) Cl2(g) Br2(l) Hg(l) Na(s)

The standard molar enthalpy of formation, HfHf for 1 mol of substance produced from elements in standard statesTabulated for compounds at 1atm and 20C

C(s) + O2(g) CO2(g) Hf = -393.5kJ/mol

Calculating H of a chemical reaction Subtract Hf values of reactants from Hf values of productsMultiply by the stoichiometric coefficient for that species.

C(s) and O2(g) Hf = 1 x 0kJ/mol = 0kJ/mol eachCO2(g) Hf = 1 x -393.5kJ/mol = -393.5kJ/mol

Hf = H products - H reactants= -393.5 - (0+0) = -393.5kJ/mol

27

Calculate H for the combustion of C2H5OH(l) Hf

values are given below

C2H5OH(l)+3 O2(g)2CO2(g) + 3H2O(l)

Hf Reactants

C2H5OH(l) = -277.7 kJ/mol x 1 mol = -277.7kJO2(g) = 0 kJ/mol x 3 mol = 0.0kJTotal Reactants: = -277.7kJ

Hf Products

CO2(g) = -393.5 kJ/mol x 2 mol = -787.0kJH2O (l) = -285.8 kJ/mol x 3 mol = -857.4kJTotal Products: =-1644.4kJ

H=products - reactants = (-1644.4kJ)-(-277.7kJ) = -1366.7kJ

28

The Indirect Method:Hess’s Law Of Constant Heat Summation

The heat of a reaction, H, is constant, whether the reaction is carried out directly in one step or through a number of steps.

The equation for the overall reaction is the sum of two or more other equations (reaction steps).

Hrxn = H1 + H2 + H3 +...

Reverse a chemical equation: Reverse the sign of H.

Hright = - Hleft

29

An Enthalpy Diagram1. Find each reactant in 1 equation only

Change equation if necessary2. Find each product in 1 equation only

Change equation if necessary3. Add reactions

Cross out product/reactant pairsAdd multiples

4. Check final equation

Find: C(graphite) + O2(g) CO(g) + 1/2O2(g) H= ?a. CO(g) + 1/2O2(g) CO2 H= -283.0kJb. C(graphite) + O2(g) CO2 H= -393.5kJ

1. C(graphite) + O2(g) CO2 H= -393.5kJ 2. CO2 CO(g) + 1/2O2(g) H= +283.0kJ 3. C(graphite) + O2(g) CO(g) + 1/2O2(g) H= -110.5kJ

30

A Styrofoam cup calorimeter was used to determine Hsol, of KOH(s).Hsol = -58.4 kJ for the reaction: KOH(s)KOH(aq)

What is Hsol,when KOH(s) reacts with H2S04 (aq)?

Use the H from the earlier calorimetry example 2 KOH(aq) + H2SO4(aq)K2SO4(aq) + 2H2O(l) H = 2 moles x –55.9 kJ/mol = -111.8 kJ

Look up H of KOH(s) to KOH(aq) and convert for 2 moles KOHKOH(s)KOH(aq) H = -58.4 kJ/molH = 2 moles x -58.4 kJ/mol = -116.8 kJ

Add the Hs for all steps of the reactions2 KOH(s) 2 KOH(aq) H = -116.8 kJ2 KOH(aq) + H2SO4(aq)K2SO4(aq) + 2H20(l) H = -111.8 kJ2 KOH (s) + H2SO4(aq)K2SO4(aq) + 2H20(l) H = -228.6 kJ